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I'm currently learning about gauge theory and am getting confused about where all the fields of a given theory take values in.

Given some $G$-gauge theory, the gauge fields take values in the adjoint representation of $G$. What representation the do matter fields take values in, such as bosonic and fermionic fields? For fermionic fields this has something to do with the spin group, but what is the connection with $G$?

Could anyone give me a brief rundown? The sources I'm looking at are mixing the Poincare group and symmetry groups and it's getting confusing, I'd like just a clear statement about where these fields lie if possible!

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Matter fields transform under the fundamental representation of the gauge group G, while the gauge fields transform under the adjoint representation of the same group. The Poincare group is different from the gauge group in the sense that it is a global symmetry group (i.e the generated transformations are not dependent on $x^\mu$) whereas gauge groups are local symmetry groups where the generated transformations are dependent on $x^\mu$.

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  • $\begingroup$ Exactly the response I wanted! Any chance you could outline the difference between fermionic and bosonic fields? $\endgroup$ – Spongebob Aug 21 '16 at 6:56
  • $\begingroup$ For the matter fields, they transform under the fundamental representation whether it is fermionic or bosonic, eg. electrons / higgs both transform under the fundamental representation. Is this the question which you asked? $\endgroup$ – Bruce Lee Aug 21 '16 at 7:00
  • $\begingroup$ Yeah kind of. Given they both transform under the same representation, yet we know that bosonic fields and fermionic fields are different, I guess I'm asking how? Is it that the fermionic fields are given by spinors, whilst bosonic are given by tensors? $\endgroup$ – Spongebob Aug 21 '16 at 7:30
  • $\begingroup$ ohh..they are different as the commutator of bosonic fields at different points vanishes, whereas the anticommutator of fermionic fields at different points vanishes. Why bosonic fields are different from fermionic fields has a different answer from why both transform under the same fundamental representation. $\endgroup$ – Bruce Lee Aug 21 '16 at 7:33
  • $\begingroup$ Fermions and bosons live under different representations of the Lorentz group, which is not a local symmetry group G. One of the possible representations of the lorentz group is spinors, which fits spin-1/2 fermions, another is scalar, which fits spin-0, vectors fit spin-1. This is totally different from the fundamental representations of the local gauge group G, under which matter fields transform. I think you are confusing both up. $\endgroup$ – Bruce Lee Aug 21 '16 at 7:40
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I disagree with Bruce Lee's answer. Matter fields can lie in any representation of the gauge group's Lie algebra - what representation they lie in needs to be specified when defining the theory. (Well, technically matter fields don't lie in the representation itself, but rather in the vector space that the representation acts on.) For example, in the Standard Model, both the Higgs field and the Weyl field $l$ containing the neutrino field and the $e$ part of the charged lepton lie in the fundamental rep of $\mathfrak{su}(2)$ but the trivial rep of $\mathfrak{su}(3)$, while the $\bar{e}$ part of the charged lepton lies in the trivial rep of both $\mathfrak{su}(2)$ and $\mathfrak{su}(3)$. Similarly, the $q$ field containing the $u$ and $d$ parts of the quarks lies in the $3$ (fundamental) rep of $\mathfrak{su}(3)$ and the fundamental rep of $\mathfrak{su}(2)$, while the $\bar{u}$ and $\bar{d}$ parts of the quarks lie in the $\bar{3}$ (complex conjugate of the fundamental) rep of $\mathfrak{su}(3)$ but the trivial rep of $\mathfrak{su}(2)$.

The gauge fields themselves lie in different representations of the gauge group's Lie algebra in different terms in the Lagrangian, depending on what they are coupled to. In the gauge fields' kinetic term $-\frac{1}{2} \text{Tr} \left( F^{\mu \nu} F_{\mu \nu} \right)$, they lie in the fundamental rep. In terms that take the form $D_\mu \varphi D^\mu \varphi$ where the gauge fields couple to matter fields via the covariant derivative, they lie in the same rep as the matter fields to which they are coupled.

Now assume the gauge field lies in the fundamental representation, and consider an infinitesimal gauge transformation $U = I - i g \theta(x) + o(\theta^2)$. The scalar field $\theta_{ij}(x)$ is a linear operator in the fundamental rep of the Lie algebra. But if we decompose it in a particular basis of generators $\{ T^a_{ij} \}$ as $\theta_{ij}(x) = \phi_a(x) T^a_{ij}$, then each component of the new field $\phi_a(x)$ specifies one generator's weight in the infinitesimal transformation. $\phi(x)$ therefore lies in the vector space acted on by the Lie algebra's adjoint representation, so mathematically it looks just like a matter field in the adjoint representation. The gauge field transforms as $A_\mu(x) \to A_\mu(x) - D_\mu \phi(x)$, where the gauge field outside the covariant derivative is in the fundamental rep, but the gauge field inside the covariant derivative lies in the adjoint rep (because it's coupled to a scalar in the adjoint rep).

Edit: upon further reflection, which representation the gauge field lies in depends on how you think about it. If you expand out the covariant derivative, then the term that couples the gauge and matter fields always looks like $$g\, \bar{\Psi}_i(x)\, \gamma^\mu\, A_\mu^a(x)\, (T_R)^a_{ij}\, \Psi_j(x),$$ where I've suppressed the spinor indices. $T_R$ represents the generators of the gauge group in the representation $R$. The index $a$ always runs over the different generators - or equivalently, it runs over the indices of the adjoint-representation vector space. The indices $i$ and $j$ run over the basis vectors of the vector space corresponding to the representation $R$. There are two different valid ways of thinking about this interaction term: you can consider the gauge field itself to just be $A_\mu^a(x)$. Under this definition, the gauge field always lies in the vector space corresponding to the adjoint representation, and the tensor $(T_R)^a_{ij}$ is like a three-point coupling between the matter fields and the gauge field. I believe that these are the more traditional words to use to describe the interaction.

But there's another way to think about it, which is the approach that Srednicki's QFT textbook uses and the way I used above: the field $A_\mu^a(x)$ is always contracted with the generator matrices $(T_R)^a_{ij}$ in some representation. (Although sometimes this contraction isn't explicit, because the generator matrices are implicitly "hidden" in the invariant symbols $f^{abc}$ or $d^{abc}$, which are defined in terms of them.) So it's arguably more straightforward to think of the contracted quantity $$(A_{R\mu})_{ij}(x) := A_{\mu}^a(x)\, (T_R)^a_{ij}$$ as the fundamental field - one fewer index to worry about. (In this way of thinking, the fields $A_{\mu}^a(x)$) are simply the components of the fundamental field $(A_\mu^R)_{ij}(x)$ in a particular basis of generators.) Under this definition, the gauge field no longer lies in the vector space of the adjoint representation, but instead it's an operator on the vector space of the arbitrary representation $R$. The two ways of thinking are completely equivalent, they only differ in which side of the definition above is considered to be the gauge field itself and which side is a quantity derived from the gauge field. But in the second way of thinking, the gauge field operators act on different representations of the gauge Lie algebra in different coupling terms of the Lagrangian, and therefore in terms where the representation is not faithful, some of its degrees of freedom get projected out. Some people find this distasteful.

Things are a bit different for the kinetic term $\mathcal{L}_\text{kin} = -\frac{1}{4} F^{a \mu \nu} F_{a \mu \nu}$ (under the first convention, where the $A_\mu^a$ color-vectors are considered the primary fields) or $\mathcal{L}_\text{kin} = -\frac{1}{2} \text{Tr}(F_{\mu \nu} F^{\mu \nu})$ (under the second convention, where the $(A_\mu)_{ij}$ color-operators are considered the primary fields). It turns out that here you don't even need to specify a representation at all - you only need the information contained in the abstract Lie algebra structure, where the elements of the Lie algebra aren't thought of as linear operators with a commutator structure, but just as vectors with an abstract Lie bracket structure. The Lie bracket structure is enough to define a symmetric, bilinear inner product on the abstract Lie algebra called the (negative of the) Killing form. When we say "Trace" in the kinetic term, we really mean this Killing form - we don't need to actually take the trace of any linear operator in any particular representation. So the kinetic term is completely representation-independent. (We typically require the gauge group to be compact and semisimple in order for the Killing form to have definite signature, so that the kinetic term is bounded below.)

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    $\begingroup$ +1 We are free to choose matter fields representations in gauge theories and that is what define a particular theory. $\endgroup$ – Diracology Feb 23 '17 at 9:56

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